Physical Quantities, the SI System and Orders of Magnitude

Physics is the business of measuring the world and then reasoning about the numbers that come back. Before a single equation of mechanics can be written, we need an agreement about what is being measured, in what units, and to what precision the answer deserves to be quoted. This chapter builds that agreement. It looks slight next to the derivations that follow it, but almost every mark lost on an IB physics paper for a "silly mistake" is lost here: a unit left off, a prefix mishandled, a result quoted to eight digits when the data supported two.

Three ideas carry the whole chapter. First, every physical quantity is a number multiplied by a unit, and the unit is part of the quantity, not decoration attached to it. Second, all units in physics are built from a tiny set of base units by multiplication and division alone, which means every equation must balance dimensionally — a fact that will let you catch algebra errors without checking the algebra. Third, a great deal of physics can be done to within a factor of ten, by order-of-magnitude estimation, long before an exact calculation is possible.

What to have to hand. A calculator that displays scientific notation, and the IB data booklet — you are permitted it in the examination, and the metric prefixes, constants and unit definitions in it are meant to be used rather than memorised.

1. Physical Quantities: Number and Unit

A physical quantity is any property of a system that can be measured and expressed as a number multiplied by a unit:

Writing $L = 2.5~\text{m}$ asserts that the length $L$ is $2.5$ times the length of the standard metre. The same length is $250~\text{cm}$ or $2.5 \times 10^{-3}~\text{km}$; the quantity has not changed, only the unit chosen to express it. This is why a bare number is meaningless in physics, and why an answer without a unit earns no marks: $9.81$ is not an acceleration, but $9.81~\dfrac{\text{m}}{\text{s}^{2}}$ is.

It also means units obey ordinary algebra. If a car travels $s = 120~\text{m}$ in $t = 8.0~\text{s}$, then

where the unit $\dfrac{\text{m}}{\text{s}}$ arose from dividing metres by seconds exactly as the number $15$ arose from dividing $120$ by $8.0$. Treat units as algebraic factors throughout a calculation and unit errors become impossible to hide.

2. The Seven SI Base Quantities

The Système International (SI) fixes seven quantities as base quantities. Every other physical quantity in the whole of physics — force, energy, charge, magnetic flux density — is derived from these by multiplication and division:

• Mass, $m$ — the kilogram, $\text{kg}$
• Length, $l$ — the metre, $\text{m}$
• Time, $t$ — the second, $\text{s}$
• Electric current, $I$ — the ampere, $\text{A}$
• Thermodynamic temperature, $T$ — the kelvin, $\text{K}$
• Amount of substance, $n$ — the mole, $\text{mol}$
• Luminous intensity, $I_v$ — the candela, $\text{cd}$

Two features of this list are worth noticing. The base unit of mass, uniquely, already carries a prefix — the base unit is the kilogram, not the gram, so $\text{kg}$ and not $\text{g}$ is what belongs in $F = ma$. And temperature in SI is measured in kelvin, an absolute scale: a temperature difference of $1~\text{K}$ equals a difference of $1~^\circ\text{C}$, but the two scales have different zeros, related by $T/\text{K} = \theta/^\circ\text{C} + 273.15$.

Since 2019 all seven units have been defined by fixing the numerical values of seven constants of nature — the second by the caesium hyperfine transition frequency, the metre by the speed of light $c = 299\,792\,458~\dfrac{\text{m}}{\text{s}}$ exactly, the kilogram by the Planck constant $h$. No physical artefact defines a unit any longer, which is why the definitions are the same in every laboratory in the universe.

Base versus derived. A base unit is one of the seven; a derived unit is any product of powers of base units. Some derived units are given their own names for convenience — the newton, the joule, the volt — but each of those is only shorthand: $1~\text{N} = 1~\dfrac{\text{kg}\cdot\text{m}}{\text{s}^{2}}$, always.

3. Derived Units and Dimensional Analysis

To express a derived unit in base units, take the defining equation of the quantity and substitute the units of everything on the right-hand side.

Worked example — the joule. Work is force times displacement, and force is mass times acceleration:

The same result follows from kinetic energy, $E_k = \tfrac{1}{2}mv^2$, whose units are $\text{kg} \cdot \left(\dfrac{\text{m}}{\text{s}}\right)^2 = \dfrac{\text{kg}\cdot\text{m}^{2}}{\text{s}^{2}}$ — and it must, because work and kinetic energy are the same kind of quantity. Two expressions for the same physical quantity can never disagree about units.

This observation is the engine of dimensional analysis: since an equation asserts that two quantities are equal, and equal quantities have equal units, every term of a physically valid equation must have the same units. A dimensionally inconsistent equation is certainly wrong; a dimensionally consistent one may still be wrong (a stray factor of $2$ or $\pi$ carries no units), but it has passed a genuine test.

Worked example — checking an equation. Is $v^2 = u^2 + 2as$ dimensionally consistent? The left side has units $\left(\dfrac{\text{m}}{\text{s}}\right)^2 = \dfrac{\text{m}^{2}}{\text{s}^{2}}$. On the right, $u^2$ matches it, and

so all three terms agree and the equation survives. Had a candidate mis-remembered it as $v^2 = u^2 + 2as^2$, the last term would carry $\dfrac{\text{m}^{3}}{\text{s}^{2}}$ and the error would be exposed in one line without any recollection of the derivation.

Dimensional analysis can also construct relationships. Suppose the period $T$ of a simple pendulum depends only on its length $l$, the mass $m$ of the bob and $g$. Writing $T \propto l^{\alpha} m^{\beta} g^{\gamma}$ and matching units,

Matching kilograms gives $\beta = 0$ — the period cannot depend on the mass. Matching seconds gives $-2\gamma = 1$, so $\gamma = -\tfrac{1}{2}$; matching metres gives $\alpha + \gamma = 0$, so $\alpha = \tfrac{1}{2}$. Hence $T \propto \sqrt{\dfrac{l}{g}}$, which is the correct pendulum law up to the dimensionless factor $2\pi$ that no dimensional argument could ever supply.

4. Scientific Notation and Metric Prefixes

Physical quantities span an absurd range — from the $10^{-15}~\text{m}$ radius of a nucleus to the $10^{26}~\text{m}$ scale of the observable universe — so numbers are written in scientific notation, as a coefficient between $1$ and $10$ multiplied by an integer power of ten:

Alternatively, a metric prefix replaces the power of ten. The ones needed in this course are:

The one trap worth rehearsing is that a prefix attached to a unit is raised to the same power as the unit. Since $1~\text{cm} = 10^{-2}~\text{m}$,

not $10^{-2}~\text{m}^3$. A density quoted as $8.9~\dfrac{\text{g}}{\text{cm}^{3}}$ therefore converts to SI as

a factor of one thousand, and the single most common conversion slip in the whole subject.

5. Significant Figures and How Precisely to Quote an Answer

The number of significant figures in a measurement states how well it is known. Reading a value of $2.50~\text{m}$ from a metre rule claims knowledge to the nearest centimetre; writing $2.5~\text{m}$ claims only the nearest $10~\text{cm}$; writing $2.5000~\text{m}$ claims a precision no metre rule possesses. The rules:

• Non-zero digits always count.
• Zeros between non-zero digits count: $1.05$ has three.
• Leading zeros never count — they only locate the decimal point: $0.0034$ has two.
• Trailing zeros after a decimal point count: $1.30$ has three, and the final zero is a claim about precision.
• Trailing zeros in an integer are ambiguous ($1500$ could be two, three or four), which is one more reason to write $1.5\times 10^{3}$ instead.

For calculations, a result can be no more precise than the data it came from:

• After multiplication or division, quote the answer to the same number of significant figures as the least precise input.
• After addition or subtraction, quote it to the same number of decimal places as the input with fewest.

So $\dfrac{4.63~\text{m}}{2.1~\text{s}} = 2.204\ldots \approx 2.2~\dfrac{\text{m}}{\text{s}}$ (two significant figures, limited by $2.1$), while $12.11~\text{m} + 3.2~\text{m} = 15.31 \approx 15.3~\text{m}$ (one decimal place). Round only at the end of a multi-step calculation; carrying intermediate values at full calculator precision avoids accumulating rounding error, and the IB mark scheme accepts answers within the tolerance this produces.

6. Orders of Magnitude and Estimation

The order of magnitude of a quantity is the power of ten nearest to it: the diameter of a hydrogen atom, $1.06\times 10^{-10}~\text{m}$, is of order $10^{-10}~\text{m}$. Comparing two quantities by order of magnitude means comparing their ratio: a proton ($10^{-15}~\text{m}$) is five orders of magnitude — a factor of $10^{5}$ — smaller than an atom, which is why the atom is overwhelmingly empty space.

A useful set of anchors:

Worked estimation. Roughly how many atoms are in a human body? Take a mass of $\sim 10^{2}~\text{kg}$, mostly water, so an average atomic mass of order $10~\text{u} \approx 10 \times 1.7\times10^{-27} \approx 2\times 10^{-26}~\text{kg}$. Then

No step of that was precise, and the answer is nonetheless right to within a factor of a few — which is exactly what an order-of-magnitude estimate promises. Such an estimate is also the fastest sanity check available: if a final answer for the mass of a car comes out as $10^{9}~\text{kg}$, something has gone wrong three lines earlier, and you know it without re-reading the algebra.

7. Common Exam Pitfalls

• Omitting the unit from a final answer. A number alone is not a physical quantity, and the mark scheme treats it as an incomplete answer even when the arithmetic is perfect.
• Forgetting that the base unit of mass is the kilogram. Substituting a mass in grams into $F = ma$ or $E_k = \tfrac{1}{2}mv^2$ produces an answer wrong by a factor of $1000$.
• Mishandling prefixes on squared or cubed units. $1~\text{km}^{2} = 10^{6}~\text{m}^{2}$ and $1~\text{cm}^{3} = 10^{-6}~\text{m}^{3}$ — the prefix is raised to the power too.
• Quoting far too many significant figures. Copying the whole calculator display implies a precision the data never had; match the least precise input instead.
• Rounding intermediate results. Round once, at the end. Rounding at every stage of a long calculation can shift the final digit and put the answer outside the accepted tolerance.
• Assuming dimensional consistency proves an equation. It only rules equations out. Dimensionless factors such as $\tfrac{1}{2}$, $2\pi$ or $\sqrt{3}$ are invisible to the method.

Summary

• Every physical quantity is a numerical value multiplied by a unit, and units follow ordinary algebra through a calculation.
• SI defines seven base quantities — mass ($\text{kg}$), length ($\text{m}$), time ($\text{s}$), current ($\text{A}$), temperature ($\text{K}$), amount ($\text{mol}$) and luminous intensity ($\text{cd}$). Every other unit is a product of powers of these.
• Derived units reduce to base units through their defining equations, e.g. $1~\text{N} = 1~\dfrac{\text{kg}\cdot\text{m}}{\text{s}^{2}}$ and $1~\text{J} = 1~\dfrac{\text{kg}\cdot\text{m}^{2}}{\text{s}^{2}}$.
• Dimensional analysis: every term of a valid equation shares the same units. It can disprove an equation and can construct proportionalities such as $T \propto \sqrt{\dfrac{l}{g}}$, but it never fixes dimensionless constants.
• Prefixes are powers of ten attached to a unit, and a prefixed unit raised to a power carries the prefix to that power: $1~\text{cm}^{3} = 10^{-6}~\text{m}^{3}$.
• Significant figures: multiplication and division keep the fewest significant figures; addition and subtraction keep the fewest decimal places; round only at the end.
• An order-of-magnitude estimate places a quantity to the nearest power of ten and is the quickest available check that a detailed answer is physically sensible.